Answer
$\sum F_A=4200N$
$\sum F_B=5700N$
Work Step by Step
The impulse-momentum theorem states that $$\sum\vec{F}\Delta t=m\vec{v}_f-m\vec{v}_0$$ $$\sum\vec{F}=\frac{m\vec{v}_f-m\vec{v}_0}{\Delta t}=\frac{m}{\Delta t}(\vec{v}_f-\vec{v}_0)$$
For car A, we have the car's mass $m=1400kg$, time $\Delta t=9s$, initial velocity $v_0=0m/s$ and final velocity $v_f=27m/s$
The net average force acting on car A is
$$\sum F_A=4200N$$
For car B, we have the car's mass $m=1900kg$, time $\Delta t=9s$, initial velocity $v_0=0m/s$ and final velocity $v_f=27m/s$
The net average force acting on car B is
$$\sum F_B=5700N$$
